If f(x) = log(sec x + tan x),then f^{prime}le | vedclass

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(D) Given $f(x) = \log(\sec x + 	an x)$.To find $f^{\prime}(x)$,we differentiate with respect to $x$ using the chain rule:$f^{\prime}(x) = \frac{d}{d

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$\sqrt{2}$

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(D) Given $f(x) = \log(\sec x + 	an x)$.To find $f^{\prime}(x)$,we differentiate with respect to $x$ using the chain rule:$f^{\prime}(x) = \frac{d}{dx} [\log(\sec x + 	an x)]$$f^{\prime}(x) = \frac{1}{\sec x + 	an x} \cdot \frac{d}{dx}(\sec x + 	an x)$$f^{\prime}(x) = \frac{1}{\sec x + 	an x} \cdot (\sec x 	an x + \sec^2 x)$Factor out $\sec x$ from the numerator:$f^{\prime}(x) = \frac{\sec x(	an x + \sec x)}{\sec x + 	an x}$$f^{\prime}(x) = \sec x$Now,evaluate at $x = \frac{\pi}{4}$:$f^{\prime}\left(\frac{\pi}{4}ight) = \sec\left(\frac{\pi}{4}ight) = \sqrt{2}$.

$I$. $f'(x) = \frac{3}{4}x^{4/3} + 9$	$II$. $f(x) = \frac{9}{28}x^{7/3} - 2$
$III$. $f(x) = \frac{9}{28}x^{7/3} + 6$	$IV$. $f'(x) = \frac{3}{4}x^{4/3} - 4$

If $f(x) = \log(\sec x + \tan x)$,then $f^{\prime}\left(\frac{\pi}{4}\right) = $

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